A convex refracting surface of radius of curvature `20 cm` separates two media of refractive indices `4//3 and 1.60`. An object is placed in the first medium `(mu = 4//3)` at a distance of `200 cm` from the refracting surface. Calculate the position of image formed.

Answer

At `234.15 cm` in denser medium

Explanation

Here, `R = 20 cm, mu_(1) = 4//3, mu_(2) = 1.60`
`u = - 200 cm, v = ?`
As refraction occurs from rarer to denser medium,
therefore, `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`- (4//3)/(-200) + (1.60)/(v) = (1.6 - 4//3)/(20)`
`(1)/(150) + (8)/(5 v) = (0.27)/(20)`
`(8)/(5 v) = (0.27)/(20) - (1)/(150) = (4.05 - 2)/(300)`
`5 v xx 2.05 = 2400`
`v = (2400)/(5 xx 2.05) = 234.15 cm`
(in denser medium)